z-transform of unit step function?
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i have this n-domain function H[n]= (0.5^n)*u[n] and i need to find z transformation of this function. i've tried this one:
>> syms n z
>> y = ((0.5)^n)*heaviside(n);
>> yz = ztrans (y,n,z)
yz =
1/(2*z - 1) + 1/2
isn't supposed to be 1/(-0.5^z-1)+1 ? is there any solution to solve this z transform?
1 Commento
Risposta accettata
Geoff Hayes
il 27 Mag 2014
I don't have the Symbolic Toolbox but I think that the provided answer is correct. If
x[n] = 0.5^n * u[n]
=> x[n] = {…,0,0,0.5^0 * 0, 0.5^1 * 1, 0.5^2 * 1,…}
then the z-transform is
sum(n=-Inf:+Inf)(0.5^n)*(z^-n)
= sum(n=0:+Inf)(0.5^n)*(z^-n) % ignore all n<0
= 0.5 + sum(n=1:+Inf)(0.5^n)*(z^-n)
= 0.5 + sum(n=1::+Inf)(0.5/z)^n
= 0.5 + (0.5/z)/(1-(0.5/z)) % assuming abs(0.5/z)<1
= 0.5 + 0.5/(z-0.5)
= 1/2 + (1/2)/(z-1/2)
= 1/(2*z - 1) + 1/2
2 Commenti
Geoff Hayes
il 27 Mag 2014
As Azzi mentions below, MATLAB's heaviside defines H(0) to be 0.5, whereas other definitions of the same function define H(0) to be 1. (See http://en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument for details.)
The formula in the third row of your table assumes that H(0)=1 and can be derived similar to above.
I think that it is fine to define u[n] with heaviside(n) as long as your are aware (and state) that heaviside(0) = 0.5.
Più risposte (2)
Amine BENZOUBIR
il 22 Ott 2019
use this commande to change the origine to 1
v=1;
sympref('HeavisideAtOrigin', v);
Azzi Abdelmalek
il 27 Mag 2014
Modificato: Azzi Abdelmalek
il 27 Mag 2014
The heaveaside function of Matlab is defined with heaviside(0) equal to 0.5
If you look at the table using another definition of heaviside (e(0)=1), you will find the z-transform of a^n is z/(z-a) . The heaviside defined in Matlab can be written as
heaviside(n)=e(n)-delta(n) (delta is Kronecker function), the z-transform is z/(z-a)-0.5
In your case replace a by 0.5
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